FACTS  AND  FORMULAE  FOR  AREA  QUESTIONS

 

 

FUNDAMENTAL CONCEPTS :

I. Results on Triangles:

1. Sum of the angles of  a triangle is 180o

2. The sum of any two sides of a triangle is greater than the third side.

3. Pythagoras Theorem : In a right - angled triangle,

Hypotenuse2=Base2+Height2

4. The line joining the mid-point of a side of a triangle to the opposite vertex is called the median.

5. The point where the three medians of a triangle meet, is called Centroid. The centroid divides each of the medians in the ratio 2 : 1.

6. In an Isosceles triangle, the altitude from the vertex bisects the base.

7. The median of a triangle divides it into two triangles of the same area.

8. The area of the triangle formed by joining the mid-points of the sides of a given triangle is one-fourth of the area of the given triangle.

 

II.Results on Quadrilaterals :


1. The diagonals of a parallelogram bisect each other

2. Each diagonal of a parallelogram divides it into two triangles of the same area.

3. The diagonals of a rectangle are equal and bisect each other.

4. The diagonals of a square are equal and bisect each other at right angles

5. The diagonals of a rhombus are unequal and bisect each other at right angles

6. A parallelogram and a rectangle on the same base and between the same parallels are equal in area.

7. Of all the parallelogram of given sides, the parallelogram which is a rectangle has the greatest area.

 

IMPORTANT FORMULAE

I. 

1. Area of a rectangle = (length x Breadth)

Length =AreaBreadth  and  Breadth=AreaLength

2. Perimeter of a rectangle = 2( length + Breadth)

 

 

II. Area of square = side2=12diagonal2 

 

III. Area of 4 walls of a room = 2(Length + Breadth) x Height

 

 

IV.

1. Area of a triangle =12×base×height

2. Area of a triangle = s(s-a)(s-b)(s-c), where a, b, c are the sides of the triangle and s=12a+b+c

3. Area of an equilateral triangle =34×side2

4. Radius of incircle of an equilateral triangle of side a=a23

5. Radius of circumcircle of an equilateral triangle of side a=a3

6. Radius of incircle of a triangle of area  and semi-perimeter s=s

 

 

V.

1. Area of a parallelogram = (Base x Height)

2. Area of a rhombus = 12×Product of diagonals

3. Area of a trapezium = 12×(sum of parallel sides)×distance between them

    

 

VI.

1. Area of a cicle = πR2, where R is the radius.

2. Circumference of a circle = 2πR.

3. Length of an arc = 2πRθ360, where θ is the central angle.

4. Area of a sector = 12arc×R=πR2θ360 

 

VII.

1. Area of a semi-circle = πR22

2. Circumference of a semi - circle = πR

Q:

Classify the triangle as a type of triangle if the sides of the triangle are 6, 12 and 13 units.

 

A) Right angled triangle B) Obtuse angled triangle
C) Acute angled triangle D) None of these
 
Answer & Explanation Answer: C) Acute angled triangle

Explanation:
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0 14284
Q:

What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad ?

A) 844 B) 814
C) 840 D) 820
 
Answer & Explanation Answer: B) 814

Explanation:

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.
Area of each tile = (41 x 41) sq.cm

Required number of tiles =1517 x 902/(41 x 41) = 814.

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8 14157
Q:

What is the area of this trapezoidal garden? (All measurements are in cm)

 

A) 60 sq.cm B) 180 sq.cm
C) 210 sq.cm D) 240 sq.cm
 
Answer & Explanation Answer: D) 240 sq.cm

Explanation:
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13 14085
Q:

A rectangular plot measuring 90 metres by 50 metres is to be enclosed by wire fencing. If the poles of the fence are kept 5 metres apart, how many poles will be needed?

A) 56m B) 65m
C) 34m D) 36m
 
Answer & Explanation Answer: A) 56m

Explanation:

Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres

Two poles will be kept 5 metres apart. Also remember that the poles will be placed along the perimeter of the rectangular plot, not in a single straight line which is very important.

 

Hence number of poles required = 280 / 5 = 56

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19 13914
Q:

A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m , then the altitude of the triangle is

A) 200m B) 300m
C) 400m D) 100m
 
Answer & Explanation Answer: A) 200m

Explanation:

Let the triangle and parallelogram have common base b, 

let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), then  

Area of triangle    = 12×b×h1 

Area of rectangle  = b*h2 

As per Given,

12*b*h1=b*h2

12*b*h1=b*100

 h1=200

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13 13820
Q:

A sector of 120 degrees, cut out from a circle, has an area of 66/7 sq cm. Find the radius of the circle.

A) 1cm B) 2cm
C) 3cm D) 4cm
 
Answer & Explanation Answer: C) 3cm

Explanation:

πr=667

 r =3

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13 13769
Q:

The area of the circle is 220 sq.m, the area of the square inscribed in a circle will be

A) 49 B) 70
C) 140 D) 150
 
Answer & Explanation Answer: C) 140

Explanation:

area of a square = 12diagonal2

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22 13586
Q:

The circumference of a circle is 66 cm. Find its radius (in cm).

 

A) 21 B) 10.5
C) 4.5 D) 9
 
Answer & Explanation Answer: B) 10.5

Explanation:
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4 13566